The locus of the middle points of the chords | vedclass

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Question

(B) Let the chord pass through the origin $(0,0)$. The equation of such a line is $y = mx$.Substituting $y = mx$ into the parabola equation $y^2 = 4ax

Vedclass · Accepted Answer

$y^2 = 2ax$

Vedclass · Answer

(B) Let the chord pass through the origin $(0,0)$. The equation of such a line is $y = mx$.Substituting $y = mx$ into the parabola equation $y^2 = 4ax$,we get $(mx)^2 = 4ax$,which simplifies to $m^2x^2 - 4ax = 0$.This gives $x(m^2x - 4a) = 0$,so the intersection points are $x_1 = 0$ and $x_2 = \frac{4a}{m^2}$.The corresponding $y$-coordinates are $y_1 = 0$ and $y_2 = m(\frac{4a}{m^2}) = \frac{4a}{m}$.The midpoint $(h, k)$ of the chord is given by $h = \frac{0 + \frac{4a}{m^2}}{2} = \frac{2a}{m^2}$ and $k = \frac{0 + \frac{4a}{m}}{2} = \frac{2a}{m}$.From $k = \frac{2a}{m}$,we have $m = \frac{2a}{k}$.Substituting this into $h = \frac{2a}{m^2}$,we get $h = \frac{2a}{(2a/k)^2} = \frac{2a \cdot k^2}{4a^2} = \frac{k^2}{2a}$.Thus,$k^2 = 2ah$. Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2ax$.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

The locus of the middle points of the chords of the parabola $y^2 = 4ax$ which pass through the origin is:

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